Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "We consider the last two-digit to see if a number is divisible by $4$. We consider the last one digit to see if a number is divisible by $5$. If $$\\overline{\\textasciitilde5A2A\\textasciitilde}$$ is divisible by $5$, the ones digit can only be $5$ or $0$. When the ones digit is $0$, the number that is formed by the last two digits, $20$, is divisible by $4$. " ]

[]

[ [ { "aoVal": "A", "content": "odd　 " } ], [ { "aoVal": "B", "content": "even　 " } ], [ { "aoVal": "C", "content": "$$1995$$ " } ], [ { "aoVal": "D", "content": "$$0$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers" ]

[ "even number $$+$$ odd number $$=$$ odd number. " ]

[]

[ [ { "aoVal": "A", "content": "$$49$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$51$$ " } ], [ { "aoVal": "D", "content": "$$59$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers" ]

[ "There are $$60$$ whole numbers from $$0$$ to $$59$$. That\\textquotesingle s $$50$$ without $$0$$ to $$9$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers" ]

[ "Based on the laws relating to the parity in addition and multiplication, either $$a$$ or $$b$$ must be $$2$$. If $$a = 2$$, then $$b = 5$$, and $$a + b = 7$$; if $$b = 2$$, then $$a = 9$$; $$9$$ is not a prime number, which doesn\\textquotesingle t match the conditions in the question. Therefore, we choose B. " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Applying the Principle of Place Value" ]

[ "$$\\rm Method$$ $$1$$: Let the hundreds, tens, and units digits of the original three-digit number be $$a$$, $$b$$, and $$c$$, respectively. We are given that $$a=c+2$$. The original three-digit number is equal to $$100a+10b+c=100(c+2)+10b+c=101c+10b+200$$. The hundreds, tens, and units digits of the reversed three-digit number are $$c$$, $$b$$, and $$a$$, respectively. This number is equal to $$100c+10b+a=100c+10b+(c+2)=101c+10b+2$$. Subtracting this expression from the expression for the original number, we get $$(101c+10b+200)-(101c+10b+2)=198$$ . Thus, the units digit in the final result is $$8$$. $$\\rm Method$$ $$2$$: The result must hold for any three-digit number with its hundreds digit being $$2$$ more than the units digit. $$301$$ is such a number. Evaluating, we get $$301-103=198$$. Thus, the units digit in the final result is $$8$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$35$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ], [ { "aoVal": "E", "content": "$$30$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders" ]

[ "There are $3\\times20+20\\div2=70$ slices of beef. $70\\div2=35$ " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$\\overline{**45}$$ " } ], [ { "aoVal": "B", "content": "$$\\overline{19*8}$$ " } ], [ { "aoVal": "C", "content": "$$\\overline{23*1}$$ " } ], [ { "aoVal": "D", "content": "$$\\overline{3*49}$$ " } ], [ { "aoVal": "E", "content": "None " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers" ]

[ "According to the ones digit of perfect squares, we can eliminate choice $B$. A perfect square ending with $5$ must be a multiple of $25$, so we can eliminate choice $A$. $${{48}^{2}}=2304$$ and $${{49}^{2}}=2401$$ so we can eliminate choice $C$. ${57}^{2}=3249$. We choose option $D$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$15^{th}$$ January $$2023$$ " } ], [ { "aoVal": "B", "content": "$$16^{th}$$ January $$2023$$ " } ], [ { "aoVal": "C", "content": "$$17^{th}$$ January $$2023$$ " } ], [ { "aoVal": "D", "content": "$$18^{th}$$ January $$2023$$ " } ], [ { "aoVal": "E", "content": "$$19^{th}$$ January $$2023$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems" ]

[ "$$16$$ days later. " ]

[]

[ [ { "aoVal": "A", "content": "$$51$$ " } ], [ { "aoVal": "B", "content": "$$54$$ " } ], [ { "aoVal": "C", "content": "$$56$$ " } ], [ { "aoVal": "D", "content": "$$61$$ " } ], [ { "aoVal": "E", "content": "$$67$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder " ]

[ "Among $1, 56, 111$, \\ldots~the smallest possible $$N$$ that satisfies the two conditions is $$56$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$17$$ " } ], [ { "aoVal": "D", "content": "$$18$$ " } ], [ { "aoVal": "E", "content": "$$20$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization" ]

[ "To make the largest possible five$$-$$digit $$N$$, you should make the number in the biggest digit as large as possible. $$120=8\\times5\\times3\\times1\\times1$$. So, the sum of digits is $$8+5+3+1+1=18$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$402$$ " } ], [ { "aoVal": "B", "content": "$$403$$ " } ], [ { "aoVal": "C", "content": "$$404$$ " } ], [ { "aoVal": "D", "content": "$$410$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders->Maximum/Minimum Problems of Division without Remainders " ]

[ "First, we need to remove all the integers which are multiples of $5$, otherwise the last digit of the products is $0$ or $5$. Hence, $403$ integers need to be removed. Next, note that the last digit of each of the products below is $6$. $1\\times2\\times3\\times4\\times6\\times7\\times8\\times9$, $11\\times12\\times13\\times14\\times16\\times17\\times18\\times19$, $\\cdots\\cdots$ $2001\\times2002\\times2003\\times2004\\times2006\\times2007\\times2008\\times2009$, and the last digit of the product $2011\\times2012\\times2013\\times2014\\times2016\\times2017\\times2018$ is $4$. Hence, we need to remove one more \"$2$\"and the last digit of the product will be $2$. So the answer is $404$. " ]

[]

[ [ { "aoVal": "A", "content": "$$34$$ " } ], [ { "aoVal": "B", "content": "$$35$$ " } ], [ { "aoVal": "C", "content": "$$30$$ " } ], [ { "aoVal": "D", "content": "$$39$$ " } ], [ { "aoVal": "E", "content": "$$40$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems" ]

[ "There are $5-1=4$ nuts left after dividing at most. Thus, Summer has $5\\times7+4=39$ nuts at most. " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers" ]

[ "Since $$5\\times 4 = 20$$, the ones digit of the given product must be $$0$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ], [ { "aoVal": "E", "content": "$$4$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "$1+2-A=0$, $A=3$. " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization" ]

[ "3 set of 2$\\times$5. " ]

[]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$11$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers" ]

[ "The ten-thousands\\textquotesingle digit plus the millions\\textquotesingle~digit of $$1234567890$$ is $$6 + 4 = 10$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$1138$$ " } ], [ { "aoVal": "B", "content": "$$1226$$ " } ], [ { "aoVal": "C", "content": "$$1324$$ " } ], [ { "aoVal": "D", "content": "$$1416$$ " } ], [ { "aoVal": "E", "content": "$$1854$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Understanding Odd and Even Numbers" ]

[ "$$1138 →1\\times1\\times3\\times8=24$$ $$1226 →1\\times2\\times2\\times6=24$$ $$1324 →1\\times3\\times2\\times4=24$$ $$1416~→1\\times4\\times1\\times6=24$$ $$1854 →1\\times8\\times5\\times4=160$$ $$1854$$ is the odd one out as the product of its digits is not $$24$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$45$$ " } ], [ { "aoVal": "D", "content": "$$75$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples" ]

[ "$$1\\times 2\\times 3\\times 4\\times 5\\times (2\\times 3)=2\\times 4\\times 2\\times (1\\times 3\\times 5\\times 3)=16\\times 45$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$9$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$15$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples" ]

[ "Since $$3\\times3 =9$$, the product of two multiples of 3 is divisible by $$9$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$23$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$29$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization" ]

[ "$35=5\\times 7$ $84=2\\times 2\\times 3\\times 7$ Because $B$ is the factor both number contains, $B=7$ Thus, $A=5$, $C=12$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "Both Odd " } ], [ { "aoVal": "B", "content": "Both Even " } ], [ { "aoVal": "C", "content": "Odd, Even " } ], [ { "aoVal": "D", "content": "Even, Odd " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers" ]

[ "Nil " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders" ]

[ "$$ 3\\times35 = 105$$; so $$106 \\div 3 =35$$ with remainder $$106-105=1$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$48$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples" ]

[ "List the prime factorization for $$4$$, $$6$$ and $$8$$ first. $$4=2^{2}$$, $$6=2\\times3$$, $$8=2^{3}$$. The largest exponent for $$2$$ is $$3$$,~~and the largest exponent for $$3$$ is $$1$$, thus the least common multiple for $$4$$, $$6$$, and $$8$$ is $$2^{3}\\times3=24$$. We choose $$\\rm C$$． " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples" ]

[ "For each one of them to get the same amount of apples, the total amount should be a multiple of $4$, because there are $4$ of them. The apples are sold in pack of six. $3$ packs contain $18$ apples; $5$ packs contain $30$; $6$ packs contain $36$; $7$ packs contain $42$. Among the number of $18, 30, 36, 42$, only $36$ can be divided by $4$. $36$ is a multiple of $4$. Therefore, they should buy $6$ packs. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$Yes.$$ " } ], [ { "aoVal": "B", "content": "$$No.$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers" ]

[ "If there are odd number of odd page number, the sum of page numbers is odd. " ]

[]

[ [ { "aoVal": "A", "content": "$$1918$$ " } ], [ { "aoVal": "B", "content": "$$1988$$ " } ], [ { "aoVal": "C", "content": "$$1998$$ " } ], [ { "aoVal": "D", "content": "$$19818$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Numbers" ]

[ "$$1000+900+80+18=1980+18=1998$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$24$$ " } ], [ { "aoVal": "B", "content": "$$36$$ " } ], [ { "aoVal": "C", "content": "$$48$$ " } ], [ { "aoVal": "D", "content": "$$72$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers" ]

[ "Let $x$ represent the length of each side of the square, then: $x^{4} = 1296 \\implies x^{2} = \\sqrt{1296} = 36$. Hence $x = \\sqrt{36} = 6$ and $4x = 4\\times 6 = 24$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$45870$$ " } ], [ { "aoVal": "B", "content": "$$45875$$ " } ], [ { "aoVal": "C", "content": "$$45579$$ " } ], [ { "aoVal": "D", "content": "$45875$ and $45870$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "The last three digits must be divisible by $125$, so it can only be $875$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$101$$ " } ], [ { "aoVal": "B", "content": "$$111$$ " } ], [ { "aoVal": "C", "content": "$$99$$ " } ], [ { "aoVal": "D", "content": "$$113$$ " } ], [ { "aoVal": "E", "content": "None of the above. " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers" ]

[ "1,3,5,7,9 -\\/-\\/-\\/-\\/-\\/-\\/-\\/-\\/-\\/-⑤ 11,13,15,17,19-\\/-\\/-\\/-\\/-\\/-\\/-⑤ 31,33,35,37,39-\\/-\\/-\\/-\\/-⑤ 51,53,55,57,59-\\/-\\/-\\/-\\/-⑤ 71,73,75,77,79-\\/-\\/-\\/-\\/-⑤ 91,93,95,97,99-\\/-\\/-\\/-⑤ $$111$$ the 31st number is 111. " ]

[]

[ [ { "aoVal": "A", "content": "（$$2$$, $$3$$, $$15$$），（$$12$$, $$30$$） " } ], [ { "aoVal": "B", "content": "（$$2$$, $$12$$, $$15$$），（$$3$$, $$30$$） " } ], [ { "aoVal": "C", "content": "（$$2$$, $$3$$, $$30$$），（$$12$$, $$15$$） " } ], [ { "aoVal": "D", "content": "（$$12$$, $$3$$, $$15$$），（$$2$$, $$30$$） " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->Finding Factors Given the Product" ]

[ "omitted " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers" ]

[ "$127$, $131$, $137$, $139$ " ]

[]

[ [ { "aoVal": "A", "content": "$$41$$ " } ], [ { "aoVal": "B", "content": "$$51$$ " } ], [ { "aoVal": "C", "content": "$$61$$ " } ], [ { "aoVal": "D", "content": "$$71$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers" ]

[ "Since $$51=3\\times17$$, $$51$$ is not a prime number. " ]

[]

[ [ { "aoVal": "A", "content": "On " } ], [ { "aoVal": "B", "content": "Off " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers" ]

[ "NA " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "$$5$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples" ]

[ "$(17+19)\\div5=7R1$ $5-1=4$ " ]

[]

[ [ { "aoVal": "A", "content": "$10\\text{cm}$ " } ], [ { "aoVal": "B", "content": "$11\\text{cm}$ " } ], [ { "aoVal": "C", "content": "$12\\text{cm}$ " } ], [ { "aoVal": "D", "content": "$13\\text{cm}$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers" ]

[ "The length of each side of the square is $\\sqrt{144} = 12\\text{cm}$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$4$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "$6\\times5-26=4$. " ]

[]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$19$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization" ]

[ "If $$1\\times2\\times3\\times4\\times5\\times6\\times7\\times$$$$8\\times9\\times10\\times11\\times12\\times13\\times14\\times15 = $$$$1307 674 368000$$, and we multiply each of these $$15$$ numbers by $$10$$, the new product will have an additional $$15$$ zeroes, and $$15+4=19$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$5786$$ " } ], [ { "aoVal": "B", "content": "$$5787$$ " } ], [ { "aoVal": "C", "content": "$$5788$$ " } ], [ { "aoVal": "D", "content": "$$5789$$ " } ], [ { "aoVal": "E", "content": "$$5784$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules" ]

[ "The number can be written as $$n+(n+1)=2n+1(n\\geqslant 1)$$ and $$x+(x+1)+(x+2)=3x+3$$. It must be a multiple of $3$ but leaves a remainder of $1$ when divided by $2$ but leaves a remainder of $1$. $5790$ can be divisible by both $2$ and $3$, so it is $5790-3=5787$. " ]

[]

[ [ { "aoVal": "A", "content": "$$99$$ " } ], [ { "aoVal": "B", "content": "$$25900$$ " } ], [ { "aoVal": "C", "content": "$$35904$$ " } ], [ { "aoVal": "D", "content": "$$34589$$ " } ], [ { "aoVal": "E", "content": "$$39804$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules" ]

[ "$99\\div3=33$ $32+33+34=99$ $32\\times33\\times34=35904$ " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder " ]

[ "The remainder of $$A$$ $$\\div9$$ is $$7$$, and $$2A=A+A$$. Therefore the remainder of $$2A\\div9$$ is $$7+7=14$$. $$14= 9+5$$, therefore the remainder is $$5$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$27$$ " } ], [ { "aoVal": "B", "content": "$$31$$ " } ], [ { "aoVal": "C", "content": "$$35$$ " } ], [ { "aoVal": "D", "content": "$$39$$ " } ], [ { "aoVal": "E", "content": "$$43$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Chinese Remainder Theorem" ]

[ "The number of Salah\\textquotesingle s cards is $$3$$ more than a multiple of $$4$$, and $$4$$ more than a multiple of $$5$$. Of the two-digit options given, the only one that satisfies both criteria is $$39(=9\\times4+3 =7\\times5+4)$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules" ]

[ "Let the original number be ab, and reverse it to be ba, calculated by the place value principle, ba-ab=72 10 + a - b (10) a + b = a = 9 b - 9, 72-8 a = b, b = 9, a = 1, the original number is 19, last digit is 9. " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers" ]

[ "The only positive whole number whose square is equal to its square root is $$1$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "The sums of the digits of $528$, $1356$, and $3336$ are multiples of $3$. " ]

[]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$32$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$40$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples" ]

[ "Keep dividing each by $$2$$ until you get an odd number. $$\\text{A}$$.$$2\\times15$$. $$\\text{B}$$.$$2\\times2\\times 2 \\times 2\\times2\\times1$$. $$\\text{C}$$.$$2\\times2\\times9$$. $$\\text{D}$$.$$2\\times2 \\times 2\\times5$$. " ]

[]

[ [ { "aoVal": "A", "content": "Possible " } ], [ { "aoVal": "B", "content": "Not possible " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers" ]

[ "For example you have number A and B. A+B = C A-B = D (A+B)+(A-B)=2A Thus, C+D must always be even number. Cannot be odd number. " ]

[]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples" ]

[ "Factors of $60$: $1$, $2$, $3$, $4$, $5$, $6$, $10$, $12$, $15$, $20$, $30$ and $60$, Factors of $75$: $1$, $3$, $5$, $15$, $25$ and $75$, Since Weili wanted to pack the apples and pears into as many bags as possible, we need to choose the largest common factor of both $60$ and $75$. Largest common factor $\\rightarrow 15$, Hence, Weili should use $15$ bags. Number of apples in each bag$\\rightarrow60\\div15$ $=4$. " ]

[]

[ [ { "aoVal": "A", "content": "$$765$$ " } ], [ { "aoVal": "B", "content": "$$675$$ " } ], [ { "aoVal": "C", "content": "$$585$$ " } ], [ { "aoVal": "D", "content": "$$495$$ " } ], [ { "aoVal": "E", "content": "$$305$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples" ]

[ "Note that $$45=5\\times 9$$. As $$5$$ and $$9$$ are coprime, a positive integer is a multiple of $$45$$ if and only if it is a multiple of both $$5$$ and $$9$$. The units digit of all five options is $$5$$, so they are all multiples of $$5$$. An integer is a multiple of $$9$$ if and only if the sum of its digits is also a multiple of $$9$$. The sums of the digits of the five options is $$18$$, $$18$$, $$18$$, $$18$$ and $$8$$. So $$305$$ is the only one of the options which is not a multiple of $$9$$ and hence is not a multiple of $$45$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$384$$ " } ], [ { "aoVal": "B", "content": "$$385$$ " } ], [ { "aoVal": "C", "content": "$$386$$ " } ], [ { "aoVal": "D", "content": "$$387$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers" ]

[ "We can list out the perfect squares: $1\\times 1 = 1$ $2\\times 2 = 4$ $3\\times 3 = 9$ $4\\times 4 = 16$ $5\\times 5 = 25$ $6\\times 6 = 36$ $7\\times 7 = 49$ $8\\times 8 = 64$ $9\\times 9 = 81$ $10\\times 10 = 100$ Hence the sum of the first $10$ perfect squares is $385$. " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ], [ { "aoVal": "E", "content": "$$1$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->The Number of Zeros at the end of a Product" ]

[ "omitted " ]

[]

[ [ { "aoVal": "A", "content": "$$13 $$ " } ], [ { "aoVal": "B", "content": "$$23 $$ " } ], [ { "aoVal": "C", "content": "$$31 $$ " } ], [ { "aoVal": "D", "content": "$$41 $$ " } ], [ { "aoVal": "E", "content": "$$43$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers" ]

[ "First note that Arthur can write down three squares, namely $$16$$, $$25$$ and $$36$$. Also, he can write down four triangular numbers, namely $$15$$, $$21$$, $$36$$ and $$45$$. If he chooses $$16$$ and $$45$$ for the square and triangular number respectively, then the remaining digits are $$2$$ and $$3$$, the prime is $$23$$. If he chooses $$25$$ and $$36$$ then the remaining digits are $$1$$ and $$4$$, the prime is $$41$$. If he chooses $$36$$ for the square number, the remaining difits can be a prime. So the largest prime he write is $$41$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$150$$ " } ], [ { "aoVal": "B", "content": "$$250$$ " } ], [ { "aoVal": "C", "content": "$$350$$ " } ], [ { "aoVal": "D", "content": "$$450$$ " } ], [ { "aoVal": "E", "content": "$$550$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers" ]

[ "There are only 2 ways for an odd number of even digits: 1 even digit or all even digits. Case 1: 1 even digit There are $5 \\cdot 5=25$ ways to choose the odd digits, 5 ways for the even digit, and 3 ways to order the even digit. So, $25 \\cdot 5 \\cdot 3=375$. However, there are $5 \\cdot 5=25$ ways that the hundred\\textquotesingle s digit is 0 and we must subtract this from 375 , leaving us with 350 ways. Case 2: all even digits There are $5 \\cdot 5 \\cdot 5=125$ ways to choose the even digits, and $5 \\cdot 5=25$ ways where the hundred\\textquotesingle s digit is 0 . So, $125-25=100$. Adding up the cases, the answer is $100+350=$ (D) 450 . " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Prime Factorization (Equations)" ]

[ "$$2016=2\\times2\\times2\\times2\\times2\\times3\\times3\\times7$$; there are $$3$$ different primes. " ]

[]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$31$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$33$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers" ]

[ "A whole number is a perfect square if it is the product of two equal whole numbers. Thus, $$1\\times1$$ and $$2\\times 2$$ and $$3\\times 3$$ and $$4\\times 4$$ are perfect squares. Continue until the product is bigger than $$1000: 30\\times 30 = 900$$; $$31\\times 31 = 961$$; $$32\\times 32 = 1024$$ (too big). " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$23$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$29$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization" ]

[ "$45=3\\times 3\\times 5$ $50=2\\times 5\\times 5$ Because $B$ is the factor both number contains, $B=5$ Thus, $A=9$, $C=10$. " ]

[]

[ [ { "aoVal": "A", "content": "$$139$$ " } ], [ { "aoVal": "B", "content": "$$140$$ " } ], [ { "aoVal": "C", "content": "$$141$$ " } ], [ { "aoVal": "D", "content": "$$142$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples" ]

[ "The number of cards after adding $$1$$ is divisible by $$4$$, $$5$$, and $$7$$. Since the least common multiple of $$4$$, $$5$$, and $$7$$ is $$4\\times5\\times7=140$$, Eddie has $$140-1=139$$ cards at least. " ]

[]

[ [ { "aoVal": "A", "content": "$$42$$ " } ], [ { "aoVal": "B", "content": "$$64$$ " } ], [ { "aoVal": "C", "content": "$$50$$ " } ], [ { "aoVal": "D", "content": "$$48$$ " } ], [ { "aoVal": "E", "content": "$$52$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules" ]

[ "Considering the ones digit and the value of the product, which is less than $10000$, only $14, 16,$ and $18$ match the condition. " ]

[]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$64$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers" ]

[ "The square root of $$16$$ is $$4$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$91$$ " } ], [ { "aoVal": "B", "content": "$$96$$ " } ], [ { "aoVal": "C", "content": "$$97$$ " } ], [ { "aoVal": "D", "content": "$$99$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers" ]

[ "We only have one prime number between $$90$$ and $$100$$, which is $$97$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ], [ { "aoVal": "E", "content": "$$36$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Common Factors and Common Multiples" ]

[ "Multiple of $$3$$ and $$4$$ also a multiple of $$12$$. $$12, 24, 36, 48, \\cdots $$ Smallest multiple between $$14$$ and $$40$$ is $$24$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$21$$ " } ], [ { "aoVal": "B", "content": "$$22$$ " } ], [ { "aoVal": "C", "content": "$$152$$ " } ], [ { "aoVal": "D", "content": "$$383$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Prime Factorization (Equations)" ]

[ "Since $$231=3\\times7\\times11$$, the sum of its prime factors is $$3 +7+11=21$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$100$$ " } ], [ { "aoVal": "D", "content": "$$255$$ " } ], [ { "aoVal": "E", "content": "$$2525$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Understanding Odd and Even Numbers" ]

[ "Let the required result be $$S$$. Then $$S= \\left( {2+4+6+\\cdots +100} \\right)- \\left( {1+3+5+\\cdots +99} \\right)~ $$ $$\\textasciitilde\\textasciitilde= \\left( {2-1} \\right) + \\left( {4-3} \\right) + \\left( {6-5} \\right) +\\cdots +\\left( {100-99} \\right) $$ $$\\textasciitilde\\textasciitilde=50\\times 1=50$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$43$$ " } ], [ { "aoVal": "B", "content": "$$33$$ " } ], [ { "aoVal": "C", "content": "$$23$$ " } ], [ { "aoVal": "D", "content": "$$13$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples" ]

[ "All but $$33$$ can be represented as required, as shown below. A. $$43=1+42$$ B. $$33$$ C. $$23=2+21$$ D. $$13=6+7$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$89$$ " } ], [ { "aoVal": "B", "content": "$$92$$ " } ], [ { "aoVal": "C", "content": "$$95$$ " } ], [ { "aoVal": "D", "content": "$$97$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers" ]

[ "Because $$99$$ is an odd number, one of these two prime numbers must be an even prime number. The only choice is $$2$$. So the other is $$99-2 = 97$$. So, the difference between the two numbers is $$95$$. Therefore, we choose $$\\rm C$$ . " ]

[]

[ [ { "aoVal": "A", "content": "$$174 $$ " } ], [ { "aoVal": "B", "content": "$$185 $$ " } ], [ { "aoVal": "C", "content": "$$198 $$ " } ], [ { "aoVal": "D", "content": "$$209 $$ " } ], [ { "aoVal": "E", "content": "$$296$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Applying Special Prime Numbers" ]

[ "The prime digits are $$2$$, $$3$$, $$5$$ and $$7$$. So the largest two-digit integer whose digits are both prime is $$77$$. However, $$77$$ is not prime, nor is $$75$$, but $$73$$ is prime. So Andy writes down $$73$$. The smallest three-digit integer whose digits are both prime is $$222$$. However, $$222$$ is not prime, but $$223$$ is prime. So Baker writes down $$223$$. Therefore the answer which Carl obtains is $$73 +223 = 296$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$1994$$ " } ], [ { "aoVal": "B", "content": "$$2042$$ " } ], [ { "aoVal": "C", "content": "$$2050$$ " } ], [ { "aoVal": "D", "content": "$$2060$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Characteristics of Remainder " ]

[ "$$497+498+499+500 =1994$$; $$509+510+511+512 = 2042$$; $$511+512+513+514=2050$$. (Divide by $$4$$, and \"start\"~near the quotient.) " ]

[]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "$$3$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$13$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$39$$ " } ], [ { "aoVal": "D", "content": "$$42$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems" ]

[ "$$3+2+2=7$$; $100-5=95$ $95\\div7=13R4$; $$13$$$\\times$$$3$$+$$3$$=$$42$$. " ]

[]

[ [ { "aoVal": "A", "content": "√√ " } ], [ { "aoVal": "B", "content": "√$$\\times $$ " } ], [ { "aoVal": "C", "content": "$$\\times $$√ " } ], [ { "aoVal": "D", "content": "$$\\times $$$$\\times $$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Casting Out Nines" ]

[ "$$(1)$$ Dividing the numbers on the left-hand side of the equals sign by $$9$$ yields remainders of $$8$$, $$4$$, and $$6$$. The sum of the remainders divided by $$9$$ is $$0$$. Dividing the right-hand side of the equals sign by $$9$$ yields a remainder of $$1$$. Therefore, $$0$$ does not equal $$1$$, making the equation incorrect. $$(2) $$The remainders on both sides of the equation are $$6$$. We can check by performing the calculation normally. It is correct. " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "We consider the last two-digit to see if a number is divisible by $4$. We consider the last one digit to see if a number is divisible by $5$. If $$\\overline{\\textasciitilde5A2A\\textasciitilde}$$ is divisible by $5$, the ones digit can only be $5$ or $0$. When the ones digit is $0$, the number that is formed by the last two digits, $20$, is divisible by $4$. " ]

[]

[ [ { "aoVal": "A", "content": "$$2013$$ " } ], [ { "aoVal": "B", "content": "$$2011$$ " } ], [ { "aoVal": "C", "content": "$$2009$$ " } ], [ { "aoVal": "D", "content": "$$2008$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders" ]

[ "For every $$4$$ girls and $$1$$ boy, the difference is $$3$$. The difference is always divisible by $$3$$. Of the choices, only $$2013$$ is divisible by $$3$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers" ]

[ "$10=2\\times 5$, the length is $5$ and the width is $2$. $(2+5)\\times 2=14$. " ]

[]

[ [ { "aoVal": "A", "content": "$$23\\times34$$ " } ], [ { "aoVal": "B", "content": "$$34\\times45$$ " } ], [ { "aoVal": "C", "content": "$$45\\times56$$ " } ], [ { "aoVal": "D", "content": "$$56\\times67$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization" ]

[ "Of the options given, $$23\\times 34$$, $$56\\times 67$$ and $$67\\times 78$$ are all not divisible by $$5$$, so may be discounted. Also $$34$$ is not divisible by $$4$$ and $$45$$ is odd, so $$34\\times 45$$ may also be discounted as it is not divisible by $$4$$. The only other option is $$45\\times 56$$. As a product of prime factors, $$45\\times 56=2^{3}\\times3^{2}\\times5\\times7$$, so it is clear that it is divisible by all of the integers from $$1$$ to $$10$$ inclusive. " ]

[]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$35$$ " } ], [ { "aoVal": "D", "content": "$$42$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples" ]

[ "$\\textasciitilde$ $\\textasciitilde$ $\\textasciitilde$ $\\textasciitilde$ " ]

[]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders" ]

[ "When $$14$$ (or any other even multiple of $$7$$) is divided by $$7$$, the remainder is $$0$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$24$$ " } ], [ { "aoVal": "C", "content": "$$28$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples" ]

[ "$3\\times 8=24$, so $24$ is a multiple of $8$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$20$$ " } ], [ { "aoVal": "B", "content": "$$23$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$29$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization" ]

[ "$35=5\\times 7$ $84=2\\times 2\\times 3\\times 7$ Because $B$ is the factor both number contains, $B=7$ Thus, $A=5$, $C=12$. " ]

[]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$11$$ " } ], [ { "aoVal": "D", "content": "$$37$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "Of the following choices, only $$37$$ is a factor of $$111$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$490$$ " } ], [ { "aoVal": "B", "content": "$$558$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$627$$ " } ], [ { "aoVal": "E", "content": "$$452$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers" ]

[ "Odd numbers end with 1, 3, 5, 7, 9 " ]

[]

[ [ { "aoVal": "A", "content": "$$43$$ " } ], [ { "aoVal": "B", "content": "$$45$$ " } ], [ { "aoVal": "C", "content": "$$47$$ " } ], [ { "aoVal": "D", "content": "$$49$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers" ]

[ "We have $$3$$ prime numbers between $$40$$ and $$50$$: $$41$$, $$43$$, $$47$$, so we choose $$\\text{C}$$. " ]

[]

[ [ { "aoVal": "A", "content": "odd　 " } ], [ { "aoVal": "B", "content": "even　 " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "prime　 " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Multiplication Rules of Odd and Even Numbers" ]

[ "Eliminate choices. Try several examples. A good example is $$7\\div4$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$64$$ " } ], [ { "aoVal": "C", "content": "$$128$$ " } ], [ { "aoVal": "D", "content": "$$256$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers" ]

[ "$$\\left(\\sqrt{64}+\\sqrt{64}\\right)^{2}=\\left(8+8\\right)^{2}=\\left(16\\right)^{2}=256$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples" ]

[ "If the product of two whole numbers is $$5$$, then one of the numbers is $$5$$ and the other is $$1$$. Their sum is $$6$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases" ]

[ "$$61725$$ " ]

[]

[ [ { "aoVal": "A", "content": "an odd number " } ], [ { "aoVal": "B", "content": "an even number " } ], [ { "aoVal": "C", "content": "a prime number " } ], [ { "aoVal": "D", "content": "a multiple of $3$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Addition and Subtraction Rules of Odd and Even Numbers" ]

[ "The sum is always odd. For example, $15 + 10 = 25$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$x\\textgreater y\\textgreater z$ " } ], [ { "aoVal": "B", "content": "$y\\textgreater z\\textgreater x$ " } ], [ { "aoVal": "C", "content": "$x\\textgreater y=z$ " } ], [ { "aoVal": "D", "content": "$z\\textgreater y\\textgreater x$ " } ], [ { "aoVal": "E", "content": "$z\\textgreater x\\textgreater y$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization" ]

[ "$x=3$, $y=2$, $z=7.$ " ]

[]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization" ]

[ "Note first that $$7632 =2\\times2\\times2\\times2\\times3\\times3\\times53$$. Therefore either the two-digit number $$de = 53$$ or the three-digit number $$abc$$ is a multiple of $$53$$. Since the multiplication uses each of the digits $$1$$ to $$9$$ once and $$7632$$ contains a $$3$$, the option $$de= 53$$ is not allowable. Hence we need to find a three-digit multiple of $$53$$ that does not share any digits with $$7632$$ and divides into $$7632$$ leaving an answer that also does not share any digits with $$7632$$. We can reject $$2 \\times 53 = 106$$ since it contains a $$6$$ but $$3 \\times 53 = 159$$ is a possibility. The value of $$7632\\div159$$ is $$2\\times2\\times2\\times2\\times3 = 48$$ which does not have any digits in common with $$7632$$ nor with $$159$$. We can also check that no other multiple of $$53$$ will work. Therefore the required multiplication is $$159 \\times 48 = 7632$$ and hence the value of $$b$$ is $$5$$. " ]

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[ [ { "aoVal": "A", "content": "$$2007$$ " } ], [ { "aoVal": "B", "content": "$$2009$$ " } ], [ { "aoVal": "C", "content": "$$2012$$ " } ], [ { "aoVal": "D", "content": "$$2015$$ " } ], [ { "aoVal": "E", "content": "$$2019$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules" ]

[ "The number can be written as $$n+(n+1)=2n+1(n\\geqslant 1)$$ and $$x+(x+1)+(x+2)=3x+3$$. It must be a multiple of $3$ but leaves a remainder of $1$ when divided by $2$. $2022$ can be divisible by both $2$ and $3$, so it is $2022-3=2019$. " ]

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[ [ { "aoVal": "A", "content": "odd$$ $$ " } ], [ { "aoVal": "B", "content": "even$$ $$ " } ], [ { "aoVal": "C", "content": "prime$$ $$ " } ], [ { "aoVal": "D", "content": "whole$$ $$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Odd and Even Numbers->Odd and Even Applications" ]

[ "Avg of $$1$$ \\& $$5$$ is $$3$$. Avg of $$1$$ and $$7$$ is $$4$$. Both are whole numbers. " ]

[]

[ [ { "aoVal": "A", "content": "$$332$$ " } ], [ { "aoVal": "B", "content": "$$339$$ " } ], [ { "aoVal": "C", "content": "$$581$$ " } ], [ { "aoVal": "D", "content": "$$585$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Remainder Problems->Questions involving Divisions with Remainders->Relationship between Dividend, Divisor, Quotient and Remainder in Division" ]

[ "$$83\\times7=581$$ $$581+4=585$$ " ]

[]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Applying the Properties of Dividing without Remainders" ]

[ "The sum of any multiple of $$100$$ and $$84$$ is divisible by $$4$$. " ]

[]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$14$$ " } ], [ { "aoVal": "E", "content": "None of the above " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Division without Remainders->Divisibility Rules" ]

[ "Since $1001$ is a multiple of $13$, $111111 = 111 \\times 1001$ is also a multiple of $13$. It follows that both $666666$ and $444444$ are both multiples of $26$. $666666XY 444444 = 66666600000000 + XY 000000 + 444444$ $\\Rightarrow XY$ must be divisible by $13$. Smallest $X+Y=1+3=4$. " ]

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[ [ { "aoVal": "A", "content": "Only $103$ " } ], [ { "aoVal": "B", "content": "$103$ and $$127$$ " } ], [ { "aoVal": "C", "content": "$103$ and $139$ " } ], [ { "aoVal": "D", "content": "$103$, $127$, and $139$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers" ]

[ "$115=5\\times23$ " ]

[]

[ [ { "aoVal": "A", "content": "$$39$$ " } ], [ { "aoVal": "B", "content": "$$50$$ " } ], [ { "aoVal": "C", "content": "$$79$$ " } ], [ { "aoVal": "D", "content": "$$87$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers" ]

[ "We note that the other options are not prime numbers because: $39 = 3 \\times 13$ $50 = 5 \\times 10$ $87 = 3 \\times 29$ Hence by the process of elimination, $79$ is a prime number. " ]

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[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Theorem of the Number of Factors of a Number->Applying Theorem of the Number of Factors of a Number Directly->The Total Number of Factors" ]

[ "$n$ is divisible by $63$. $63=3^{2}\\times7$. $63$ has $(2+1)\\times(1+1)=6$ divisors. Thus, $n$ has at least $6$ divisors, and $n$ can\\textquotesingle t have an odd number of divisors. Option $D$ is the only possible answer. " ]

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[ [ { "aoVal": "A", "content": "$$51$$ " } ], [ { "aoVal": "B", "content": "$$52$$ " } ], [ { "aoVal": "C", "content": "$$53$$ " } ], [ { "aoVal": "D", "content": "$$59$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Factors and Multiples->Basic Concepts of Factors and Multiples" ]

[ "A prime number is a number greater than $$1$$ whose only whole number factors are itself and $$1$$. Of the numbers greater than $$50$$, $$51 = 3\\times17$$ and $$52 = 2\\times26$$. The first prime is $$53$$. " ]

[ "其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$8$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules" ]

[ "Since both numbers are divisible by $3$ , the sum of their digits has to be divisible by three. $7+4+5+2+1=19$. To be a multiple of $3$ , $A+B$ has to be either $2$ or $5$ or $8 \\ldots$ and so on. We add up the numerical digits in the second number; $3+2+6+4=15$. We then add two of the selected values, $5$ to $15$ , to get $20$ . We then see that $C=1,4$ or $7,10 \\ldots$ and so on, otherwise the number will not be divisible by three. We then add 8 to 15 , to get 23 , which shows us that $C=1$ or $4$ or $7 \\cdots$ and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be $1,4$, and $7$ . However, in the answer choices, there is no $7$ or $4$ or anything greater than $7$ , but there is a $1$ , so $(\\mathbf{A}) 1$ is our answer. " ]

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[ [ { "aoVal": "A", "content": "$$43$$ " } ], [ { "aoVal": "B", "content": "$$45$$ " } ], [ { "aoVal": "C", "content": "$$47$$ " } ], [ { "aoVal": "D", "content": "$$49$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime and Composite Numbers->Determining Prime and Composite Numbers" ]

[ "We have $$3$$ prime numbers between $$40$$ and $$50$$: $$41$$, $$43$$, $$47$$, so we choose $$\\text{C}$$. " ]

[]

[ [ { "aoVal": "A", "content": "$\\left (8231\\right )\\_5$ " } ], [ { "aoVal": "B", "content": "$\\left (2001\\right )\\_5$ " } ], [ { "aoVal": "C", "content": "$\\left (4341\\right )\\_7$ " } ], [ { "aoVal": "D", "content": "$\\left (2345\\right )\\_5$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Place Value and Number Bases->Properties and Applications of Number Bases" ]

[ "In quinary, the base number must be $5$, and all the digits in the brackets must be less than $5$. " ]

[]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ], [ { "aoVal": "E", "content": "$$1$$ " } ] ]

[ "Overseas Competition->Knowledge Point->Number Theory Modules->Prime Factorization->Applying Prime Factorization->The Number of Zeros at the end of a Product" ]

[ "omitted " ]